Find Zeros of Polynomial Function

Section 2.5
Question 56




Can you do 56 as a guide for me to do the rest on my own?

 

56.

f(x)=2x^3+3x^2+18x+27

given zero: 3i -> so we will also have -3i

factors are (x-3i) and (x+3i)

their product is: (x-3i) (x+3i)=x^2-(3i)^2=x^2-9(-1)=x^2+9

use long division
..........(2x+3
x^2+9| 2x^3+3x^2+18x+27
..........2x^3+0*x^2+18x
......................3x^2+27
.........................3x^2+27
...................................0

last factor is 2x+3 and third zero is
2x+3=0
2x=-3
x=-3/2

so, zeros are: 3i, -3i, -3/2

 

In terms of long division, what term is the dividend and which one is the divisor?

 

The number that is being divided (in this case, 2x^3+3x^2+18x+27) is called the dividend,
and the number that it is being divided by (in this case, x^2+9) is called the divisor.

 

I could not make out the difference between the dividend and divisor as typed in your first reply.

 

I write it similar to this


I do it this way

.............(
x^2+9 | 2x^3+3x^2+18x+27

 

I now see what you're talking about. Cool. Taking a little break from math for the rest of this week. Need to rest my brain cells.

 

ok

 

I enjoy solving math problems but I need a short break. Be back Friday, Lord willing, with more questions from Section 2.6.